Answer:
The Milky Way is about 1,000,000,000,000,000,000 km (about 100,000 light years or about 30 kpc) across. The Sun does not lie near the center of our Galaxy. It lies about 8 kpc from the center on what is known as the Orion Arm of the Milky Way
I’m pretty sure u have it right
I'm actually going ahead in the book (DC Circuits) so this isn't really homework but I figured the tag was appropriate....the name of the chapter is Ohm's Law and Watt's Law.
<span>Problem: Calculate the power dissipated in the load resistor, R, for each of the circuits.Circuit (a): V = 10V; I = 100mA; R = ?; Since I know
V and
I use formula
P = IV: P = IV = (100mA)(10V) = 1 W.</span>
The next question is what I'm not sure about:
Question: What is the power in the circuit (a) above if the voltage is doubled? (Hint: Consider the effect on current).
What I did initially was: P = IV = (100mA)(2V) = 2 W
But then I looked at the answer and it said 4 W, then I looked at the Hint again. Then I remembered in the book early on it said "If the voltage increases across a resistor, current will increase."
So question is: When solving problems I have to increase (or decrease) current (I) every time voltage (V) is increased (decreased) in a problem, right? How about the other way around, when increasing current (I), you need to increase voltage (V). I'm pretty sure that's how they got 4 W, but want to make sure before I head to the next section of the book.
P = IV = (200mA)(2V) = 4 W
The universe has trillions of galaxies and counting. Astronomers give names to each galaxy base on its shape (e.i, Sombrero galaxy, Milkyway Galazy, ect,.).
Also, the size of the galaxy is taken into account and their color.
The frictional force of an object is the product of the normal force and coefficient of kinetic friction. Here the frictional force acting on the object is 16.4 N.
<h3>What is frictional force?</h3>
Frictional force is a kind of force acting on a body to resist it from motion. Thus, the direction of the force will be in negative with the magnitude. Frictional force is the product of coefficient of friction and the normal force.
The normal force acting on the object of mass 4.2 Kg is N = mg
N = 4.2 Kg × 9.8 m/s² = 41.16 N
Frictional force = ц N
= 0.40 × 41.16 N
= 16.4 N.
Therefore, the frictional force acting between the surface of the object and the floor is 16.4 N
To find more on frictional force, refer here:
brainly.com/question/1714663
#SPJ1
Your question is incomplete. But your complete question probably was:
The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40 the weight of the object is 4.2 kg. What is the frictional force of the object?
